A model jet is fired up in the air from a 16-foot platform with an initial upward velocity of 52 feet per second. The height of the jet above the ground after t seconds is given by the equation h=-16t^2+52t+16 where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot

A) 777 feet

B) 58 feet

C) 30 feet

D) 16 feet

Question

Stich

Mathematics

25 June, 21:44

A model jet is fired up in the air from a 16-foot platform with an initial upward velocity of 52 feet per second. The height of the jet above the ground after t seconds is given by the equation h=-16t^2+52t+16 where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot

A) 777 feet

B) 58 feet

C) 30 feet

D) 16 feet

+3

Answers (1)

Know the Answer?

Escobar · Answer 1

B. 58 feet to nearest foot.

Step-by-step explanation:

h = - 16t^2 + 52t + 16

Convert to vertex form:-

= - 16 [ (t^2 - 3.25t) + 16

= - 16 [ (t - 1.625) ^2 - (1.625) ^2 ] + 16

= - 16 (t - 1.625) ^2 + 42.25 + 16

= - 16 (t - 1.625) ^2 + 58.25.

The maximum height occurs when t = 1.625 seconds and its equal to 58 feet.